Answer
The solution set is $\left\{-6, -1\right\}$.
Work Step by Step
RECALL:
A quadratic trinomial $x^2+bx+c$ may be factored if there are integers $d$ and $e$ such that $de = c$ and $d+e =b$
If such integers exist, then
$x^2+bx + c = (x+d)(x+e)$
The trinomial in the given equation has $c=6$ and $b= 7$.
Note that $6=6(1)$ and $6+1 = 7$
This means that $d=6$ and $e = 1$ and the given trinomial may be factored as:
$v^2+7v+6
\\=(v+d)(v+e)
\\=(v+6(v+1)$
Thus, the given equation is equivalent to:
$(v+6)(v+1) =0$
Use the Zero-Product Property by equating each factor to zero then solving each equation to obtain:
$\begin{array}{ccc}
&v+6=0 &\text{ or } &v+1=0
\\&v=-6 &\text{ or } &v=-1
\end{array}$
Therefore the solution set is $\left\{-6, -1\right\}$.
